AcWing 837. 连通块中点的数量

连通块中点的数量 - 并查集

#include <iostream>

using namespace std;

const int N = 1e6 + 10;

int p[N], cnt[N];
int n, m;
int find(int x){
	if(p[x] != x) p[x] = find(p[x]);
	return p[x];
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++){
		p[i] = i;
		cnt[i] = 1;
	}
	while(m--){
		char op[5];
		int a, b;
		scanf("%s", op);
		if(op[0] == 'C'){
			scanf("%d%d", &a, &b);
			if(find(a) == find(b)) continue;
			cnt[find(b)] += cnt[find(a)];
			p[find(a)] = find(b);
		}
		else if(op[1] == '1'){
			scanf("%d%d", &a, &b);
			if(find(a) == find(b))puts("Yes");
			else puts("No");
		}
		else{
			scanf("%d", &a);
			printf("%d\n", cnt[find(a)]);
		}
	}
	return 0;
}

#include <iostream>

using namespace std;

const int N = 1e6 + 10;

int n, m;
int p[N], si[N];//不能用size关键字定义 C++中

int find(int x){
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main(){
    scanf("%d%d", &n, &m);
    
    for(int i = 1; i <= n; i++){
        p[i] = i;
        si[i] = 1;
    }
    
    while(m--){
        char op[5];
        int a, b;
        scanf("%s", op);
        
        if(op[0] == 'C'){
            scanf("%d%d", &a, &b);
            if(find(a) == find(b)) continue;//如果本来就在一个集合内 就跳过
            si[find(b)] += si[find(a)];
            p[find(a)] = find(b);
        }
        else if(op[1] == '1'){
            scanf("%d%d", &a, &b);
            if(find(a) == find(b)) puts("Yes");
            else puts("No");
        }
        else{
            scanf("%d", &a);
            printf("%d\n", si[find(a)]);
        }
    }
    
    return 0;
}